For #f(x) = 2x^2 - 4x -1#, what is the vertex and axis of symmetry?

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Answer 1 · 2017-12-09T20:22:41

vertex at #(1,-3)#, the axis of symmetry is at #x=1#

There are two ways of doing this - completing the square or through differentiation. Either way, we are trying to find the minimum point.

Completing the square

#f(x)=2x^2-4x-1#
#=2(x^2-2x-1/2)#
#=2([x-1]^2-1-1/2)#
#=2([x-1]^2-3/2)#
#=2[x-1]^2-3#

This means the minimum value is at #f(x)=-3#, where #x=1#.

The axis of symmetry will be the same as the x-coordinate of the minimum, this being #x=1#

So the vertex is at #(1,-3)#, the axis of symmetry is at #x=1#

Differentiation

I'd generally only differentiate if the number were a bit awkward, if completing the square would give loads of fractions.

#f(x)=2x^2-4x-1#
#f'(x)=4x-4#
Let #f'(x)=0#
#4x-4=0#
#x=1#
Let #x=1#
#f(1)=2(1)^2-4(1)-1#
#=-3#

So the vertex is at #(1,-3)#, the axis of symmetry is at #x=1#